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  1. The value of diamond varies directly as the square of its weight. If a diamond falls and breaks into two pieces with weights in the ratio 2:3. what is the loss percentage in the value?Sol:Let weight be “x”
    the cost of diamond in the original state is proportional to x2when it is fallen it breaks into two pieces 2y and the 3y
    x = 5y
    Original value of diamond = (5y)2 = 25y2
    Value of diamond after breakage = (2y)2+(3y)2=13y2

    so the percentage loss will be = 25y2−13y225y2×100=48%

    2. Five college students met at a party and exchanged gossips. Uma said, “Only one of us is lying”. David said, “Exactly two of us are lying”. Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are lying”. Chitra said “All of us are lying”. Which one was telling the truth?


    Sol:  As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.
    Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.

    3. Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that.

    a) 2676
    b) 2
    c) 445
    d) 86

    SOL: This is a number series problem nothing to do with the data given.
    1x 1+1=2
    2 x 2+2=6
    6 x 3+3=21
    21 x 4+4=88 and not 86
    88 x 5+5 = 445
    445*6+6 = 2676

    4. The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?

    a) AOTDSP
    b) AOTPDS
    c) AOTDPS
    d) AOSTPD

    In alphabetical order : A D O P S T
    A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120.  So the word starts with A.
    A D _ _ _ _ : empty places can be filled in 4!=24
    A O _ _ _ _ : the places filled with 4! ways = 24.  If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
    A O D _ _ _ : 3!= 6
    A O P _ _ _ : 3!=6
    Till this 36 words are obtained, we need the 42nd word.
    AOS _ _ _ : 3!= 6
    Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
    So given word is AOSTPD

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4. A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?

SOL:  This is a case of without replacement.  We have to multiply two probabilities.  1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.

5. There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?
Sol: At least 3 reds means we get either : 3 red  or  4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21

Total probability =  = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
= 312/16807

6. Total number of 4 digit number do not having the digit 3 or 6.

consider 4 digits  _ _ _ _
1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0)
2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9)
3st blank can be filled in 8C1 ways
4st blank can be filled in 8C1  ways
Therefore total 4 digit number without 3 or 6 is 7 x 8 x 8 x 8=3584

7. Find the missing in the series:  70, 54, 45, 41,____.

Sol: 40
70-54 = 16 = 42
54-45 = 9 = 32
45-41 = 4 = 22
41-40 = 1 = 12

8. A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and also all rooms have equal number of students. What is the least number of rooms needed?

Sol:  We have to find the maximum number which divides all the given numbers so that number of roots get minimized.  HCF of 120,192 & 144 is 24.  Each room have 24 students of the same course.
Then rooms needed 12024+19224+14424  = 5 +8 + 6 = 19

9. A farmer has a rose garden. Every day he picks either 7,6,24 or 23 roses. When he plucks these number of flowers the next day 37,36,9 or 18 new flowers bloom. On Monday he counts 189 roses. If he continues on his plan each day, after some days what can be the number of roses left behind? (Hint : Consider number of roses remaining every day)

let us consider the case of 23. when he picks up 23 roses the next day there will be 18 new, so in this case., 5 flowers will be less every day. So when he counts 189, the next day 184, 179,174,169,…………….
finally the no. of roses left behind will be 4.

10. What is the 32nd word of “WAITING” in a dictionary?

Sol: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W

Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways
so can’t be arranged starting with A alone as it is asking for 32nd word so it is out of range

AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways.  Out of range as it has to be within 32 words.
AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24
AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways
so,24+12 =36th word so out of range. So we should not consider all the words start with AGN
now AGNI_ _ _can be arranged in 3! ways =6 ways
so 24+6=30 within range
Now only two word left so, arrange in alphabetical order.
AGNTIIW  – 31st word
AGNTIWI  – 32nd word

11. A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10. How many “different” boxes of chocolates can be made?

If n similar articles are to be distributed to r persons, x1+x2+x3……xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1
In this case x1+x2+x3……x6=10
in such a case the formula for non negative integral solutions is n+r−1Cr−1
Here n =6 and r=10. So total ways are 10+6−1C6−1 = 3003

12. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.
a. 1/3
b. 1/2
c. 5/9
d. 17/36

Sol: Their sum can be 3,4,6,8,9,12
For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 – n) ways.
Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
So probability is (20/36)=(5/9)

13. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work?
a. 2000
b. 4000
c. 5000
d. 3000

Time taken by A and B is in the ratio of =  3:2
Ratio of  the Work = 2 : 3 (since, time and work are inversely proportional)
Total money is divided in the ratio of 2 : 3 and B gets Rs.3000

14. On a 26 question test, 5 points were deducted for each wrong answer and 8 points were added for right answers. If all the questions were answered how many were correct if the score was zero.
a. 10
b. 11
c. 13
d. 12

Let x ques were correct. Therefore, (26- x) were wrong
Solving we get x=10

15. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be,
a. 5.5
b. 4.5
c. 7.5
d. 6.5

Volume =l×b×h = 6×5×2  = 60 cm3
Now volume is reduced by 19%.
Therefore, new volume = (100−19)100×60=48.6
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume: (x100×6)(x100×5)2=48.6
Solving we get x =90
thus length and width is reduced by 10%
New width = 5-(10% of 5)=4.5

16. If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used?

Sol: We have to consider the number of 4’s in two digit numbers. _ _
If we fix 4 in the 10th place, unit place be filled with 10 ways.  If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed)
So total 19 ways.
There are total 9 4’s in 14, 24, 34…,94
& total 10 4’s in 40,41,42….49
thus, 9+10=19.

17. If twenty four men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve men and thirty seven women work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work?

Sol: Let man daily wages and woman daily wages be M and W respectively
solving the above equations gives M=350 and W=200

18. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow?

Profit = 4200
Profit =SP – CP
4200=SP – 300000 therefore SP=304200
x+y = 300000
1.2x + 0.9y = 304200
Solving for x = 114000 = CP of cow.

19. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4……
In the above sequence what is the number of the position 2888 of the sequence.

a) 1
b) 4
c) 3
d) 2

Sol: First if we count 1223334444. they are 10
In the next term they are 20
Next they are 30 and so on
So Using n(n+1)2×10≤2888

For n = 23 we get LHS as 2760.  Remaining terms 128.

Now in the 24th term, we have 24 1’s, and next 48 terms are 2’s.  So next 72 terms are 3’s.
The 2888 term will be “3”.

20. How many 4-digit numbers contain no.2?

Sol: Total number of four digit numbers =9000 (i.e 1000 to 9999 )
We try to find  the number of numbers not having digit 2 in them.
Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here ‘0’ cannot be taken
Total number of numbers not having digit 2 in it =9 x 9  x 9 x 8 =5832
Total number of numbers having digit 2 in it = 9000-5832 =3168

1) A power unit is there by the bank of a river of 900 meters. A cable is made from power unit to power a plant opposite to that of the river of 2000mts. The cost of the cable below water is Rs. 5/- per meter and cost of cable on the bank is Rs. 4/- per meter. Then find out the amount to be invested to connect those two stations?




Required length of wire = 2000 mts cost of cable below water = 900 * 5 = 4500 cost of cable on the bank of river= (2000 – 900) * 4 = 4400 Total cost = 4500 + 4400 = 8900.


2) A farmer had 20 hens. All but 2 died. How many hens are still alive?





“All but 2” means 2. So, all but 2 died means that there are only two hens are alive and others were dead. So, the answer is 2.


3) At this series: 7, 10, 8, 11, 9, 12… What number should come next?




This is a simple alternating addition and subtraction series. In the first pattern, 3 are added; in the second, 2 are subtracted.


4) A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?





We can select the 5 member team out of the 10 in 10C5 ways = 252 ways. The captain can be selected from amongst the remaining 5 players in 5 ways. Therefore, total ways the selection of 5 players and a captain can be made = 252*5 = 1260.


5) There are 3 gentlemen in a meeting: Mr. Yellow, Mr. Green and Mr. Brown. They are wearing yellow, green and brown ties. Mr. Yellow says: “Did you notice that the colors of our ties are different from our names?” The person who is wearing the green tie says, “Yes, you are right!”  Do you know Brown is wearing what colour of tie?





We know that Mr. Yellow was not wearing a yellow tie because of his statement. He also was not wearing the green tie because the one wearing the green tie agreed to his statement. Therefore, Mr. Yellow was wearing a brown tie. Mr. Green was wearing a yellow tie. And Mr. Brown was wearing the green tie.


6) g[0]=1,g[1]=-1,g[n]=2*g[n-1]-3*g[n-2] then calculate g[4]= ?





From the given function g[n] =2*g [n-1]-3*g [n-2] put values of n as first 2 then 3, then 4 and you will get the answer. G[2] = 2*g[1]-3*g[0]=-2-3 =-5 , G[3]=2*g[2]-3*g[1] =2*(-5)-3*(-1)=-7, G [4] =2*g[3]-3*g[2]=2*(-7)-3*(-5) =1


7) Ten years ago X was half of Y. If the ratio of their present ages is 3:4, what will be the total of their present ages?




Let X’s age 10 years ago = x years. Then, Y’s age 10 years ago = 2x years. (x + 10)/ (2x + 10) = 3/4 => 4 (x + 10) = 3 (2x + 10) x = 5 Total of their present ages = (x + 10 + 2x + 10) = 3x + 20 = 15 + 20 = 35 years.


8) A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:


Answer: 120KMPH


9) In a circular racetrack of length 100 m, three persons A, B and C start together. A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively. While C runs in the opposite at 15 m/s. When will all the three meet for the first time after the start?





Since the track is a circular track A and B will meet every 50 seconds. i.e. 100 / (10-8). Since it is a multiple of 50 they will be meeting at the starting point every 50 Seconds. If you multiply 15 x 50 you will get 750 and after the second 50 it will be 1500. All of them will meet at the starting point after 100s


10) A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?





Let the base be X. Therefore (X*X + X*0 + 0) = (2*X +4) + (3*X + 2), X * X = 5 * X + 6, X * X – 5 * X -6 = 0, (X- 6) (X+1) = 0 Therefore base is 6.


11) Siva, Sathish, Amar and Praveen are playing cards. Amar is to the right of Sathish, who is to the right of Siva. Who is to the right of Amar?


Answer: Praveen


12) Ferrari is leading car manufacturer. Ferrari S.P.A. is an Italian sports car. It has enjoyed great success. If Mohan’s Ferrari is 3 times faster than his old MERCEDES which gave him 35 km/ph if  Mohan travelled 490 km in his Ferrari the how much time in hours he took?




Time = (Distance/Speed) = (490/35) Therefore the total time = time/3 = (490/35)/3 = 4.6666 Approximately 4.7


13) My flight takes off at 2 AM from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N 70W. What is the local time when my plane landed?




Total change in degrees in east-west direction = 70 + 10 degrees =80 degrees, Time change due to the change in Longitude = 80*4 = 320 minutes = 5 hrs 20 min. According to the person the time would have been = 2 am + 10 hrs = 12 am. But according to the local time the time would be = 12 – 5.20 = 6 .40am


14) A class of 25 students took a science test. 10 students had an average (arithmetic mean) score of 80. The other students had an average score of 60. What is the average score of the whole class?




80 × 10 + 60 × 15 = 800 + 900 = 1700 Total number of students = 25 Average weight = 1700/25 The average score of the whole class is 68.


15) The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens and two erasers is Rs.32. How much will three pencils, three pens and three erasers cost?




x + 2y + 4z = 22 ——–>1 5x + 4y + 2z = 32 ——->2 multiply eqn (2) by 2 and substract from eqn(1) dividing by 3 we get 3x + 2y = 14———>3 multiplying eqn (1) by 5 and subtracting eqn (2) then divide by 6 we get y + 3z = 13———->4 Adding eqn (3) and (4) we get 3x + 3y +3z = 27


16) A is 20 percent more efficient than B. If the two persons can complete a piece of work in 60 days. In how many days A working alone can complete the work?





Ratio of time taken by A and B is 1: 1.2 i.e. 5:6 Let the time taken by A be 5x and by B be 6x , 1/5x + 1/ 6x = 1/60 => x =22. Time taken by A = 5 * 22 = 110


17) On a certain island, 5% of the 10000 inhabitants are one-legged and half of the others go barefooted. What is the least number of Shoes needed in the island?




In any case the one legged people will all require one shoe per head. From the remaining, half will go barefooted and therefore they need no shoes. And the rest will need two shoes per head. And this works out at one shoe per person for the others. Therefore for the whole population on the average one shoe per head


18) Look at this series: 14, 28, 20, 40, 32, 64, what number should come next?





This is an alternating multiplication and subtracting series: First, multiply by 2 and then subtract 8.


19) Two pipes A and B can fill a tank in 24 min. and 32 min. respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 minutes?




Let B be closed after x minutes. Then, part filled by (A + B) in x min. + part filled by an in (18 – x) min. = 1. X {1/24 + 1/32] + (18 – x) X 1/24 = 1 or 7x/96 + (18 – x)/24 = 1 7x + 4(18 -x) = 96 or x=8 Hence, B must be closed after 8 minutes


20) At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remaining are Lorries. If every vehicle is equally likely to leave, find the probability of car leaving second if either a lorry or van had left first :





Let S be the sample space and A be the event of a van leaving first. n(S) = 100 n (A) = 30 Probability of a van leaving first: P (A) = 30/100 = 3/10 Let B be the event of a lorry leaving first. n (B) = 100 – 60 – 30 = 10 Probability of a lorry leaving first: P (B) = 10/100 = 1/10 If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars. Let T be the sample space and C be the event of a car leaving. n (T) = 99 n(C) = 60 Probability of a car leaving after a lorry or van has left: P(C) = 60/99 = 20/33


21) A red house is built from red bricks. A blue house is built from blue bricks. A white house is built from white bricks. Then what is a greenhouse made from?





Glass i.e. Buildings are made of bricks. But a greenhouse is made of glass.


22) If Rs. 782 be divided into three parts, proportional to 1/2: 2/3: 3/4, then the first part is:  www.allindiajobs.in





Given ratio = 1/2: 2/3:3/4 = 6: 8: 9 1st part = 782*(6/22) = 213.2727.. Approximately equals to 213 www.allindiaobs.in


23) A man decides to buy a nice horse. He pays $60 for it, and he is very content with the strong animal. After a year, the value of the horse has increased to $70 and he decides to sell the horse . But after a few days later he regrets his decision to sell the beautiful horse, and he buys it again. Unfortunately he has to pay $80 to get it back, so he loses $10. After another year of owing the horse, he finally decides to sell the horse for $90. What is the overall profit the man makes?





The total expenses are $60 + $80 = $140 and the total earnings are $70 +$90 = $160. The overall profit = $160 – $140 = $ 20


24) When I become as old as my father is now, I will be 5 times the age of my son. In addition, my son will be 8 yrs older than What I am now. If the sum of my father’s age and my age is equal to100. How old is my son now?




Let us consider the ages of the man, his father & his son is m, f & s respectively. Then according to the problem F=5s………… (i) S + f-m = m+8…….. (ii) F + m = 100……….. (iii) Putting (i) in (iii) we get 5s+m=100…….. (iv) Putting (i) in (ii) we get 6s-2m=8………… (v) Equating (iv) & (v) we get s = 13, f = 65 & m = 36


25) Nine persons went to the hostel for taking their meals, 8 of them spent Rs.12 each on their meals and ninth spend Rs. 8 more then the average expenditure of all nine. What is the total money spent by them?




Let average expenditure be x Then 8*12 +(x+8) = 9x i.e. 104 + x = 9x; 8x = 104 x = 13 Therefore total money spent is 9x = 9*13 =117


26) The average of a list of 6 numbers is 20. If we remove one of the numbers, the average of the remaining numbers is 15. What is the number that was removed?




The removed number could be obtained by difference between the sum of original 6 numbers and the sum of remaining 5 numbers i.e. sum of original 6 numbers – sum of remaining 5 numbers sum of original 6 numbers = 20 × 6 = 120 sum of remaining 5 numbers = 15 × 5 = 75 Number removed = sum of original 6 numbers – sum of remaining 5 numbers 120 – 75 = 45


27) In the word ORGANISATIONAL, if the first and second, third and fourth, fifth and sixth letters are interchanged, what would be the 12th letter from right?





After interchanging R O A G I N S A T I O N A L 14 13 12 11 10 9 8 7 6 5 4 3 2 1


28) What are the number of edges, number of vertices and number of faces of a planar cube among the following options?





29) We know cube has 6 faces, 12 edges and 8 vertices


How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?




Let the number of required alphabets in the language be ‘n’. Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials. These n3 different initials = 1 million i.e. n3 = 106 (1 million = 106) => n3 = (102)3 => n = 102 = 100


30) Today’s high school students spend too much time thinking about trivial and distracting matters such as fashion. Additionally, they often dress inappropriately on school grounds. Rather than spending time writing another detailed dress policy, we should make school uniforms mandatory. If students were required to wear uniforms, it would increase a sense of community and harmony in our schools and it would instill a sense of discipline in our students. Another positive effect would be that teachers and administrators would no longer have to act as clothing police, freeing them up to focus on more important issues.


This paragraph best supports the statement that




  1. a) inappropriate clothing leads to failing grades
  2. b) students who wear school uniforms get into better colleges
  3. c) teachers and administrators spend at least 25% of their time enforcing the dress code.
  4. d) school uniforms should be compulsory for high school students



Answer: Explanation:


The support for choice e is in the third sentence “we should make school uniforms mandatory”. There is no evidence provided to support choices a, b, and d. And although we know that teachers and administrators are spending some of their time enforcing dress code, the paragraph does not quantify how much of their time is spent that way, so there is no support for choice c.


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